A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1. An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by:
E0r(r+r1).lL
LE0rlr1
LE0r(r+r1)l
E0lL
Solution:
The current in the circuit ,I=E0/(r+r1) The potential drop across the wire is V=Ir=E0r/(r+r1) Potential gradient,k=V/L=E0r/(r+r1)L Thus, emf E=kl=E0rl/(r+r1)L